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Concept:
\(Gravitaional\;Force = F \propto \frac{{{m_1}{m_2}}}{{{r^2}}} \Rightarrow F = G\frac{{{m_1}{m_2}}}{{{r^2}}}\) Where G is gravitational constant, r is the distance between two masses, m1 and m2 are the masses. Calculation: Given, F force when the distance between two bodies is d \(F = G\frac{{{m_1}{m_2}}}{{{d^2}}}\) …..Eq-1 Fine the distance at which force become 64 F Let the distance = D \(64\;F = G\frac{{{m_1}{m_2}}}{{{D^2}}}\) ……Eq-2 Dividing Eq-1 with Eq-2 We get \(\frac{F}{{64F}} = \frac{{{D^2}}}{{{d^2}}}\) \( \Rightarrow {\left( {\frac{D}{d}} \right)^2} = \frac{1}{{64}} \Rightarrow D = \frac{d}{8}\) Hence the correct option is d/8. India’s #1 Learning Platform Start Complete Exam Preparation
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Mock Tests & Quizzes Trusted by 3.3 Crore+ Students Text Solution Solution : The force of attraction between two bodies of masses `m_(1) and m_(2)` and separated by distance r is given by Newton's universal law of gravitation `i.e.," "F=(Gm_(1)m_(2))/(r^(2))` where, G is the universal constant in nature. <br> This force is know as gravitational force. The gracitational force is directly proportional to the product of the masses of two bodies and inversely proportional to the square of the distance between them. <br> All bodies fall with the same acceleration due to gravity whatever their masses have. <br> `i.e.," "g=(GM)/(R^(2))` <br> (where M = Mass of the earth, = Radius of the earth) <br> And from the equation it is clear that, acceleration due to gravity (i.e., g) depends only on mass of the earth, the radius of the earth. So, two bricks tied together will not fall faster than a single bricks under the action of gravity. Hence, the hypothesis is not correct. |
What is the relation ship between the force of attraction and the distance between two bodies
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